Lemma 68.6.4. Let $S$ be a scheme. Let $f : U \to X$ be a surjective, étale, and separated morphism of algebraic spaces over $S$. For $p \geq 0$ set

$W_ p = U \times _ X \ldots \times _ X U \setminus \text{all diagonals}$

where the fibre product has $p + 1$ factors. There is a free action of $S_{p + 1}$ on $W_ p$ over $X$ and

$\mathop{\mathrm{Hom}}\nolimits (K^ p, \mathcal{F}) = S_{p + 1}\text{-anti-invariant elements of } \mathcal{F}(W_ p)$

functorially in $\mathcal{F}$ where $K^ p = \wedge ^{p + 1}f_!\underline{\mathbf{Z}}$.

Proof. Because $U \to X$ is separated the diagonal $U \to U \times _ X U$ is a closed immersion. Since $U \to X$ is étale the diagonal $U \to U \times _ X U$ is an open immersion, see Morphisms of Spaces, Lemmas 66.39.10 and 66.38.9. Hence $W_ p$ is an open and closed subspace of $U^{p + 1} = U \times _ X \ldots \times _ X U$. The action of $S_{p + 1}$ on $W_ p$ is free as we've thrown out the fixed points of the action. By Lemma 68.6.1 we see that

$(f_!\underline{\mathbf{Z}})^{\otimes p + 1} = f^{p + 1}_!\underline{\mathbf{Z}} = (W_ p \to X)_!\underline{\mathbf{Z}} \oplus Rest$

where $f^{p + 1} : U^{p + 1} \to X$ is the structure morphism. Looking at stalks over a geometric point $\overline{x}$ of $X$ we see that

$\left( \bigoplus \nolimits _{\overline{u} \mapsto \overline{x}} \mathbf{Z} \right)^{\otimes p + 1} \longrightarrow (W_ p \to X)_!\underline{\mathbf{Z}}_{\overline{x}}$

is the quotient whose kernel is generated by all tensors $1_{\overline{u}_0} \otimes \ldots \otimes 1_{\overline{u}_ p}$ where $\overline{u}_ i = \overline{u}_ j$ for some $i \not= j$. Thus the quotient map

$(f_!\underline{\mathbf{Z}})^{\otimes p + 1} \longrightarrow \wedge ^{p + 1}f_!\underline{\mathbf{Z}}$

factors through $(W_ p \to X)_!\underline{\mathbf{Z}}$, i.e., we get

$(f_!\underline{\mathbf{Z}})^{\otimes p + 1} \longrightarrow (W_ p \to X)_!\underline{\mathbf{Z}} \longrightarrow \wedge ^{p + 1}f_!\underline{\mathbf{Z}}$

This already proves that $\mathop{\mathrm{Hom}}\nolimits (K^ p, \mathcal{F})$ is (functorially) a subgroup of

$\mathop{\mathrm{Hom}}\nolimits ((W_ p \to X)_!\underline{\mathbf{Z}}, \mathcal{F}) = \mathcal{F}(W_ p)$

To identify it with the $S_{p + 1}$-anti-invariants we have to prove that the surjection $(W_ p \to X)_!\underline{\mathbf{Z}} \to \wedge ^{p + 1}f_!\underline{\mathbf{Z}}$ is the maximal $S_{p + 1}$-anti-invariant quotient. In other words, we have to show that $\wedge ^{p + 1}f_!\underline{\mathbf{Z}}$ is the quotient of $(W_ p \to X)_!\underline{\mathbf{Z}}$ by the subsheaf generated by the local sections $s - \text{sign}(\sigma )\sigma (s)$ where $s$ is a local section of $(W_ p \to X)_!\underline{\mathbf{Z}}$. This can be checked on the stalks, where it is clear. $\square$

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